What is the significance of Terminal temperature difference (TTD) in feed water heaters ???

 

 

 

 

 

 

 

 

 

The Terminal Temperature Difference (ΔT or delta T) is a crucial concept in the field of heat transfer and thermodynamics. It represents the temperature difference between the hot and cold fluids in a heat exchanger at the point where they leave the heat exchanger. The significance of terminal temperature difference lies in its impact on the efficiency and performance of heat exchange processes.

 

Terminal temperature difference is the difference between the saturation temperature at the operating pressure of the inlet steam to the heater and the temperature of the feed water leaving the heater.

The heating steam temperature they are talking about is the saturation temperature of the steam for the given supply pressure. The highest pressure heater is almost always receiving steam that is still superheated (Energy above saturation temp/pressure) If the incoming steam has 15 degrees of superheat, and the outgoing feed water absorbs all of that and is heated to 2 degrees above the saturation temperature of the supplied heating steam pressure, You have a negative 2 degree TTD.

Most of the other heaters are heated with steam from further down the turbine steam path, and have very little or no super heat in their steam, therefore, no negative TTD.

 

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Terminal temperature difference provides feedback on the feed water heater’s performance relative to heat transfer

 Significance of TTD.

 

1-It gives the feed back on performance of heat exchanger

2-Higher TTD is nothing but thee is more difference between saturation temperature of steam and feed water leaving the heater.This indicates the poor performance of heater.Similarly lower TTD is nothing but thee is small difference between saturation temperature of steam and feed water leaving the heater.This indicates the good heat transfer between steam and feed water & hence there is better performance of heater

3-The concept of "temperature approach" is closely related to ΔT. The temperature approach is the difference between the temperature of the hot fluid and the temperature of the cold fluid at the end of the heat exchanger. A smaller temperature approach is often desired to maximize heat transfer efficiency, but it is limited by practical considerations.

4-the terminal temperature difference is a key parameter in the analysis, design, and optimization of heat exchange systems. It plays a vital role in determining heat transfer rates, efficiency, and the size of heat exchangers, ultimately impacting the performance and cost of thermal systems in various engineering applications

5-Station heat rate will improve

6-Cycle efficiency will increase

7-Less steam consumption for feed water heating

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 Calculation of TTD of feed water heater.

 1-A HP heater is used to heat the feed water from 125 °C to 160 °C by using MP steam at pressure 13 kg/cm2 at temperature 280 °C, calculate the TTD.

We have,

TTD = Saturation temperature of inlet steam - Feed water outlet temperature

Saturation temperature of inlet steam at 13 kg/cm2g pressure = 195.6 °C

TTD = 195.6 - 160 = 35.6 °C

Note: For best performance, heaters are designed to get TTD 3 to 5 °C at full operation capacity.

 2-A LP heater is used to heat the feed water from 80 °C to 110 °C by using LP steam at pressure 2.5 kg/cm2A at temperature150 °C, calculate the TTD.

 We have,

TTD = Saturation temperature of inlet steam - Feed water outlet temperature

Saturation temperature of inlet steam at 2.5 kg/cm2g pressure = 125°C

 TTD = 125-110= 15 °C

 3-A HP heater is used to heat the feed water from 105°C to 140 °C by using MP steam at pressure 8 kg/cm2 at temperature 220 °C, calculate the TTD.

We have,

TTD = Saturation temperature of inlet steam - Feed water outlet temperature

Saturation temperature of inlet steam at 8 kg/cm2g pressure = 170 °C

TTD = 170 - 140 = 30 °C

 Note: For best performance, heaters are designed to get TTD 3 to 5 °C at full operation capacity.

 4-A LP heater is used to heat the feed water from 47 °C to 78 °C by using LP steam at pressure 0.62 kg/cm2A at temperature 87 °C, calculate the TTD.

 We have,

TTD = Saturation temperature of inlet steam - Feed water outlet temperature

Saturation temperature of inlet steam at 0.62 kg/cm2g pressure = 87 °C

TTD = 87 - 78 = 9 °C

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How do you calculate the quantity of condensate generated in steam lines???

 How do you calculate the quantity condensate generated in steam lines???

 

How does condensation happens in steam line?

Condensation formation in steam lines is a common issue in steam distribution systems, and it can lead to various problems, including

• Reduced energy efficiency,
• Equipment damage,
• Operational issues.
• Poor steam quality
• Disturbances in process

How does condensation happens in steam line?

Condensation occurs when hot steam comes into contact with a surface that is cooler than its dew point temperature, causing the steam to lose heat and change phase into water droplets. Here are some factors to consider when dealing with condensation in steam lines:

Calculation:

A 100 TPH Boiler operating at of working pressure 87 kg/cm2 and 515 deg C supplies steam to 20 MW Turbine.The pressure and temperature at Turbine inlet are 85 kg/cm2 and 505 deg C, calculate the quantity of condensate formed.

 

Solution:

Enthalpy of steam at 87 kg/cm2 and 515 deg C =819 kcal/kg

Enthalpy of steam at 85 kg/cm2 and 505 deg C =814 kcal/kg

Enthalpy difference = 819-814 = 5 kcal/kg

Enthalpy of evaporation at average steam pressure 86 kg/cm2 is =332 kcal/kg

There fore,quantity of condensate generated = (100 X 5 / 332) =1.5 TPH

 

A process is situated at 500 meter from Turbine exhaust line.The exhaust pressure is 3 kg/cm2 and 150 deg C temperature and the steam paameters at process are 2.2 kg/cme and 138 deg C, quantity of steam supplied for process is 75 TPH.Calculate the condensation formed in steam line

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Solution:

Enthalpy of steam at 3 kg/cm2 and 150 deg C =657 kcal/kg

Enthalpy of steam at 2.2 kg/cm2 and 138 deg C =654 kcal/kg

Enthalpy difference = 657-654 = 3 kcal/kg

Enthalpy of evaporation at average steam pressure 2.6 kg/cm2 is =519 kcal/kg

There fore,quantity of condensate generated = (75 X 3 / 519) =0.43 TPH

 

What are the factors to be considered when dealing with steam line and condensation

 

How do you reduce steam condensation?

 

Temperature Differential: The primary cause of condensation is the temperature difference between the steam and the surrounding environment. To minimize condensation, you can either insulate the steam lines to maintain the steam's temperature or increase the temperature of the surrounding environment.

 

Insulation: Proper insulation of steam lines is crucial. High-quality insulation helps to maintain the temperature of the steam and prevents it from coming into contact with cooler surfaces. Insulation materials like fiberglass, mineral wool, or foam are commonly used for this purpose.

 

Steam Traps: Steam traps are essential components in steam systems. They are used to remove condensate from the steam lines while allowing steam to pass. Regular maintenance and inspection of steam traps can prevent condensate buildup.

 

Proper Sloping: Steam lines should be installed with a slight downward slope in the direction of condensate flow. This helps the condensate to drain away from the steam-carrying pipe, reducing the chances of condensate buildup.

 

Drainage Points: Install drainage points at low spots in the steam lines or at points where condensation is likely to occur. These drainage points should be equipped with proper traps and drains to remove condensate effectively.

 

Steam Pressure: Maintaining the proper steam pressure in the lines can also help reduce condensation. Lowering the pressure can reduce the temperature differential, which decreases the likelihood of condensation.

 

Steam Quality: Ensure that the steam quality is high. Wet or low-quality steam is more likely to condense. Proper steam generation and water treatment are essential to achieve high-quality steam.

 

Air Venting: Properly vent steam lines to remove air, which can contribute to temperature variations and condensation issues.

 

Monitoring and Maintenance: Regularly monitor steam lines for signs of condensation, such as water droplets or corrosion. Perform routine maintenance to address any issues promptly.

 

Heat Tracing: In some cases, heat tracing systems can be used to maintain the temperature of the steam lines, preventing condensation.

 

Pipe Material: The choice of pipe material can also impact condensation. Some materials, like copper, conduct heat more effectively and may be less prone to condensation compared to others.

 

Addressing condensation in steam lines is essential for the efficient and safe operation of steam systems. It helps prevent damage to equipment, ensures consistent steam quality, and reduces energy losses. Proper design, insulation, and maintenance are key to minimizing condensation-related issues in steam lines.

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Topping cycle & calculations

  Topping cycle & calculations

 

A Co-generation system can be classified as either a topping cycle or a bottoming cycle on the basis of sequence of energy generated & use.

 In a topping cycle, the fuel supplied is used to first produce power and then thermal energy, which is the by-product of the cycle and is used to satisfy process heat requirements.

 In a topping cycle, a primary heat source, such as a gas turbine or an internal combustion engine, is used to drive a generator and produce electricity. The primary cycle typically operates at higher temperatures and generates high-pressure and high-temperature exhaust gases.

 The exhaust gases from the topping cycle are then directed to a waste heat recovery boiler or a heat exchanger, where their residual heat is captured. This waste heat is then used to produce steam, which drives a steam turbine or an organic Rankine cycle (ORC) turbine in the bottoming cycle.

 Topping cycles are commonly used in combined cycle power plants, where they offer improved efficiency and performance compared to standalone gas turbines or internal combustion engines. The integration of a bottoming cycle allows for the utilization of waste heat, maximizing the overall energy output of the system.

 In a bottoming cycle, the primary fuel used produces high temperature thermal energy and the heat rejected from the process is used to generate power through a heat recovery Boiler & Turbo generator.

 Bottoming cycles are suitable for manufacturing processes that require heat at high temperature in furnaces & kiln and reject heat at significantly high temperatures.

 The bottoming cycle operates at lower temperatures and utilizes the waste heat energy to generate additional power. By extracting energy from the waste heat, the topping cycle achieves higher overall efficiency compared to a single-cycle power generation system.

 Topping cycle calculation:

 A Co-generation facility is defined as one, which simultaneously produces two or more forms of useful energy such as electrical power and steam, electric power and shaft (mechanical) power, etc.” The project may qualify to be termed as a co-generation project, if it is in accordance with the definition and also meets the qualifying requirement outlined below:

 Topping cycle mode of co-generation – Any facility that uses non-fossil fuel input for the power generation and also utilizes the thermal energy generated for useful heat applications in other industrial activities simultaneously.

 For the co-generation facility to qualify under topping cycle mode, the sum of useful power output and one half the useful thermal output be greater than 45% of the facility’s energy consumption, during season.”

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Following inputs required for calculation of topping cycle:

• Fuel consumption
• Fuel GCV
• Steam given to processes & their heat content
• Power generation

Topping cycle is calculated by using following formula

 TC Eff = (Sum of total heat supplied to process in kcal X 50% + Total electricity generated in kcal) X 100 / Fuel energy

Example

 A 44 MW Co-generation plant is operating at 41 MW load and utilizing bleed & extraction steam for process heating. Calculate the topping cycle efficiency

The inputs required are as below

Sl No	Particular/Parameters	UOM	Value
1	Boiler fuel consumption	TPH	85
2	Fuel GCV	Kcal/kg	2250
3	Process-1 steam flow	TPH	12
4	Process steam-1 enthalpy	Kcal/kg	740
5	Process-2 steam flow	TPH	170
6	Process steam-2 enthalpy	Kcal/kg	653
7	Power generation	MWH	41

 Calculation:

Total heat content in input fuel = 85 X 1000 X 2250 =191250000 kcal

Heat content in process-1 steam = 12 X 1000 X 740 =8880000 kcal

Heat content in process-2 steam = 170 X 1000 X 653 =111010000 kcal

Power generation in kcal = 41 X 1000 X 860 = 35260000 kcal

 TC Eff = (Sum of total heat supplied to process in kcal X 50% + Total electricity generated in kcal) X 100 / Fuel energy

TC Eff = ((8880000+111010000) X 50% + 35260000) X 100 / 191250000

 TC eff = 49.78%

   

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Calculation of pressure drop in steam and water lines

 Calculation of pressure drop in steam and water lines

 The pressure drop in a water & steam lines refers to the decrease in pressure that occurs as water/steam flows through a pipe or conduit due to factors such as friction and flow resistance. Several factors influence the magnitude of pressure drop in a water line:

Pipe Characteristics: The diameter, length, and roughness of the pipe impact the resistance to flow and consequently the pressure drop. Smaller diameter pipes and longer pipe lengths tend to result in higher pressure drops. Additionally, rougher pipe surfaces create more friction and increase pressure drop compared to smoother surfaces.

Flow Rate: The rate at which water/steam flows through the pipe affects the pressure drop. Higher flow rates generally result in higher pressure drops due to increased frictional resistance.

Fluid Properties: The physical properties of the water/steam being transported, such as viscosity and density, can influence the pressure drop. However, for water at typical temperatures and pressures, these effects are usually negligible.

 Pipe Fittings and Valves: The presence of fittings, such as elbows, bends, valves, and other obstructions in the water line, can contribute to pressure drop. These components disrupt the flow and introduce additional resistance.

 It's important to note that pressure drop calculations for steam lines can be complex and require a comprehensive understanding of steam properties and fluid dynamics.

 Pressure drop in water line:

Head loss in water line for turbulent flow is given as

Head loss in meter = 4fLV² / (2gD)

 Where, f = Friction loss in pipe, generally varies from 0.005 to 0.007

L = Pipe length

D = Diameter of the pipe

g = Acceleration due to gravity, 9.81 m/s2

V = Velocity of the fluid

 Example:

A Boiler feed pump is delivering feed water flow 50 TPH to the boiler at a distance of 70 meter.The steam drum height is 38 meter from pump suction.Calculate the pressure drop in water line, assume pipe line size is 80 NB, water density 980 kg/m3 & neglect the other losses from pipe line fittings.

 Feed water flow in m3/sec = 50 000 kg/hr / 980 kg/m3 = 51.02 m3/hr =0.014 m3/sec

Area in side the pipe line = 3.142 X 0.08²/4 = 0.05 M²

Feed water velocity,V =  Flow / Area = 0.014 / 0.005 =2.78 m/sec

 Then, head loss, H = 4 X 0.005 X 38 X 2.78² / (2 X 9.81 X 0.08)

Head loss, H = 3.75 meter

 Minimum head required to lift the water up to steam drum, considering pressure drop in feed water control valve is 8 kg/cm2

 H = 3.75+80+38 =121.75 meter

 Pressure drop in steam line:

Head loss in meter = 2fLdV² / (500gD)

(Density of water is 500 times more than steam at atmospheric pressure)

Where, f = Friction loss in pipe, generally varies from 0.005 to 0.007

L = Pipe length

D = Diameter of the pipe

g = Acceleration due to gravity, 9.81 m/s2

V = Velocity of the fluid

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 Example: Turbine inlet steam flow is 100 TPH & the distance between Boiler MSSV & Turbine MSSV is 82 meter.The seam pressure & temperature are 65 kg/cm2 and 490 deg C respectively.Calculate the pressure drop in steam line.

 Density of steam at above parameters = 17 kg/m3

Steam flow in m3/sec = 100 X 1000 kg/hr / 17 kg/m3=5882.35 m3/hr = 1.63 m3/sec

Assume main steam velocity being 45 m/sec

Pipe inside area A = Flow / Velocity = 1.63/45 =0.0362 m2

Now, calculate pipe diameter , A = 3.142 X D²/4

D = SQRT (0.036 X 4/3.142) = 0.214 meter = 214 mm

Now, Pressure drop H =(2 X 0.005 X 82/0.224) X (17/500) X (45²/9.81) =25.7 m = 2.57 kg/cm2

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